Question

A coin is tossed $5$ times. Find the probability of getting at least $4$ heads

Answer 1

Finding the probability of getting at least $4$ heads:

Number of times coin tossed, n $=5$

Number of possibilities $=2$ (H, T)

Number of possible outcomes $\mathrm{n}\left(\mathrm{S}\right)=2^{\mathrm{n}}=2^5=32$

P(All heads) $={\left(\frac{1}{2}\right)}^5=\frac{1}{32}$

P(at least $4$ heads)$=\mathrm{p}\left(\mathrm{x}=4\right)+\mathrm{p}(\mathrm{x}=5)$

$\begin{array}{rcl}\mathrm{P}(\mathrm{at}\mathrm{least}4\mathrm{heads})& =& \mathrm{p}\left(\mathrm{x}=4\right)+\mathrm{p}(\mathrm{x}=5)\\ & \Rightarrow & ={}^5\mathrm{C}_4{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^4+{}^5\mathrm{C}_5{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^5\\ & \Rightarrow & =\frac{5!}{4!\left(5-4\right)!}{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^4+\frac{5!}{5!\left(5-5\right)!}{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^5\\ & \Rightarrow & =\frac{5!}{4!\left(1!\right)}{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^4+\frac{5!}{5!\left(0\right)!}{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^5\\ & \Rightarrow & =\frac{5!}{4!}{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^4+\frac{5!}{5!}{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^5\\ & \Rightarrow & =\frac{5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1}\left(\frac{1}{2}\right){\left(\frac{1}{2}\right)}^4+\frac{5\times 4\times 3\times 2\times 1}{5\times 4\times 3\times 2\times 1}{\left(\frac{1}{2}\right)}^5\\ & \Rightarrow & =\left(5\times \frac{1}{2}\times \frac{1}{16}\right)+\left(\frac{1}{32}\right)\\ & \Rightarrow & =\frac{5}{32}+\frac{1}{32}\\ & \Rightarrow & =\frac{6}{32}\\ & \Rightarrow & =\frac{3}{16}\end{array}$

Hence, the probability of getting at least $4$ heads is $\frac{3}{16}$.